Why is electric flux through any closed surface $q/\epsilon_0$? In schools we are only taught of its simplest case, i.e. flux through a sphere with charge centered at origin. And then it is generalised to all closed surfaces. Is there really any proof of flux through all closed surfaces.


A simple derivation would be this one. Suppose you have a single charge $q_i$ inside a closed surface.

The flux across a closed surface is: $\Phi=\oint\vec{E} d\vec{A}$

Coulomb's law says: $\vec E=\frac{q}{4\pi\epsilon_0r^3}\vec r$

In this case, it means that the flux is: $\Phi_i =\oint \frac{q_i}{4\pi\epsilon_0r^3}\vec r d\vec{A}$

If you solve the integral, it yields: $\Phi_i=\frac {q_i}{ \epsilon_0}$

By the superpositon principle: $\Phi=\frac {Q}{ \epsilon_0}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.